Is there a formula to work out the distance from lens to object for macro ?

I have done some experiments with a 210mm lens but things don't seem to add up as I would expect

with the bellows at 420mm for 1:1 lens to film plane the object is 410mm from the lens when in focus
if I go to 630mm for 1:2 then the object is at 310mm
at 840mm for 1:4 then the object is at 275mm
at 1050mm for 1:5 the object is at 250mm
and at 1260mm for 1:6 the object is at 240mm

I,m not quite sure where on the lens I should measure from but I used the same place for all measurements which was the plane of the iris which seemed to work out very well with the 1:1 measurement

thanks

robin

C. D. Keth

20-Apr-2014, 09:18

Google the "thin lens equation."

Francisco J. Fernández

20-Apr-2014, 09:41

Hi

Is there a formula to work out the distance from lens to object for macro ?

I have done some experiments with a 210mm lens but things don't seem to add up as I would expect

with the bellows at 420mm for 1:1 lens to film plane the object is 210mm from the lens when in focus
if I go to 630mm for 1:2 then the object is at 310mm
at 840mm for 1:4 then the object is at 275mm
at 1050mm for 1:5 the object is at 250mm
and at 1260mm for 1:6 the object is at 240mm

I,m not quite sure where on the lens I should measure from but I used the same place for all measurements which was the plane of the iris which seemed to work out very well with the 1:1 measurement

thanks

robin

Hello.

According to the theory, when an image is full scale (1X) or (1:1) there are always 4 times the focal length between the focal plane and the image plane.

Only symmetric lenses without modification of design, the conjugate distances will be equal and symmetric in both sides of the optical center of the lens used.

But, if you use a telephoto lens or a lens with symmetrical design variants. The conjugates distances inside and outside the camera will no longer be symmetrical and move (with tele will lower the internal conjugate distance, the bellows will be less long than expected).

to work in the right way in macro (1X), you should place the focus plane and the image plane to 4 times the focal length used. And then focus. Then reality and its image will always have the same scale.

I am no friend of calculators, i only use the formulas only if I need escales intermediate.

In ancient lens models was very common to use symmetric designs PLANAR type (or variants) in expensive lenses. (Symmar, Xenotar)

And asymmetrical designs Tessar type (or variants) in cheaper lenses, or with the specific purpose of the portrait were used (Xenar, and I think Heliar too).

In more modern optical there are many variants (due to patent laws)

and that produces changes in the theoretical hypothesis due to the design. But the basics almost everything is as it was.

Francisco J. Fernández

20-Apr-2014, 09:58

When you see a lens with both parties on each side of the plug of the same size. You may suspect that this lens will be symmetrical and may work as I said.

In ancient Symmar (and similar) that was so. But it was not so in the convertible Symmar (its inner part is smaller and cramped).

Most discussions neglect internodal distance, which is usually small. Few of us know exactly where our lenses' nodes are. For most of the lenses used in LF photography, the nodes are close to the diaphragm; this is not the case with telephoto lenses.

Most discussions neglect internodal distance, which is usually small. Few of us know exactly where our lenses' nodes are. For most of the lenses used in LF photography, the nodes are close to the diaphragm; this is not the case with telephoto lenses.

Most discussions neglect internodal distance, which is usually small. Few of us know exactly where our lenses' nodes are. For most of the lenses used in LF photography, the nodes are close to the diaphragm; this is not the case with telephoto lenses.